ĐKXĐ: \(x\ge0\)
Ta có:
x+x+1+\(2\sqrt{x\left(x+1\right)}\)=x+2
<=>\(2\sqrt{x\left(x+1\right)}=1-x\)
<=>4x(x+1)=1-x2
<=>4x2+4x+x2-1=0
<=>5x2+4x-1=0
<=>5(x+1)(x-0,2)=0
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-0,2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\left(loại\right)\\x=0,2\left(chọn\right)\end{matrix}\right.\)
Vậy x= 0,2